Ratio of neutron power to fusion power

For PFRC parameters: Te = 30 keV, Td = 70 keV, and T(he3) ranges from 70 to 140 keV. The tritium burn-up factor is nominally 0.5. The results are consistent with Figure 3 in Santarius.

%--------------------------------------------------------------------------
% See Also: PowerDHe3
%--------------------------------------------------------------------------

%--------------------------------------------------------------------------
% Reference: Santarius, J.F. and B.G. Logan, "Generic Magnetic Fusion
%            Rocket", UWFDM-914, University of Wisconsin,  February 1998.
%--------------------------------------------------------------------------

% Tritium burnup of 50%
d5 = PowerDHe3;
d5.tD = 70;
d5.tE = 30;
d5.fT = 0.5;
d5.nD = 1e20;

% Tritium burn-up of 0 (all exhausted before it can fuse)
d0 = d5;
d0.fT = 0;

The = linspace(70,140);

ratio0 = [];
ratio5 = [];

% Ratio D-3He of 2:1
d5.nHe3 = 0.5e20;
d0.nHe3 = d5.nHe3;
for k = 1:length(The)
  d5.tHe3 = The(k);
  [pF, pN] = PowerDHe3( d5 );
  ratio5(1,k) = pN/pF;
  d0.tHe3 = The(k);
  [pF, pN] = PowerDHe3( d0 );
  ratio0(1,k) = pN/pF;
end

% Equal ratio D-3He
d5.nHe3 = 1e20;
d0.nHe3 = d5.nHe3;
for k = 1:length(The)
  d5.tHe3 = The(k);
  [pF, pN] = PowerDHe3( d5 );
  ratio5(2,k) = pN/pF;
  d0.tHe3 = The(k);
  [pF, pN] = PowerDHe3( d0 );
  ratio0(2,k) = pN/pF;
end

% Ratio D-3He of 1:3
d5.nHe3 = 3e20;
d0.nHe3 = d5.nHe3;
for k = 1:length(The)
  d5.tHe3 = The(k);
  [pF, pN] = PowerDHe3( d5 );
  ratio5(3,k) = pN/pF;
  d0.tHe3 = The(k);
  [pF, pN] = PowerDHe3( d0 );
  ratio0(3,k) = pN/pF;
end

Plot2D(The,ratio5,'Helium-3 Temperature (keV)','Ratio (Pneutron/Pfusion)','PFRC Neutron Power Fraction')
legend('3He:D=0.5','3He:D=1','3He:D=3')
hold on
colors = get(gca,'colororder');
plot(The,ratio0(1,:),'--','color',colors(1,:))
plot(The,ratio0(2,:),'--','color',colors(2,:))
plot(The,ratio0(3,:),'--','color',colors(3,:))

text(110, 0.04, 'Tritium Burnup: 0.5')
text(90, 0.016, 'Tritium Burnup: 0')

% All D-D
d5.nD = 1e20;
d5.nHe3 = 0;
[pF, pN] = PowerDHe3( d5 );
ratioDD = pN/pF;

disp('For an ion temperature of 70 keV and tritium burnup of 0.5:')
fprintf('\t3He/D of 0.5: %f\n',ratio5(1,1))
fprintf('\t3He/D of 1: %f\n',ratio5(2,1))
fprintf('\t3He/D of 3: %f\n',ratio5(3,1))
fprintf('\tD-D:\t %f\n',ratioDD)
disp('(Compare against figure in Santarius)')


%--------------------------------------

For an ion temperature of 70 keV and tritium burnup of 0.5:
	3He/D of 0.5: 0.070019
	3He/D of 1: 0.037197
	3He/D of 3: 0.012938
	D-D:	 0.595197
(Compare against figure in Santarius)

Published with MATLAB® R2020a